The best way to solve this is by using
Bunuel's method. The next best way is to use smart numbers and plug them into the various equations.
However, if you don't see that
Bunuel's method is possible during the ~ 2min you have to solve the problem or if your smart numbers aren't working, then the following method is a pretty standard way to arrive at the right solution. Keep in mind though that this method does take longer and should be used as a kind of last resort.
\(\lvert a - b\rvert > \lvert a + b\rvert\)
\(\lvert a - b\rvert = a - b\), when \(a - b > 0\)
\(\lvert a - b\rvert = -(a - b)\), when \(a - b < 0\)
\(\lvert a + b\rvert = a + b\), when \(a + b > 0\)
\(\lvert a + b\rvert = -(a + b)\), when \(a + b < 0\)
Case 1:if \(a - b > 0, a + b > 0\), then
\(a - b > a + b\)
\(0 > 2b\)
\(b < 0\)
Also, if we add \(a + b > 0\) and \(a - b > 0\) we get:
\(a > 0\)
Case 2:if \(a - b > 0, a + b < 0\), then
\(a - b > -(a + b)\)
\(a - b > -a - b\)
\(2a > 0\)
\(a > 0\)
Also, if we subtract \(a + b < 0\) from \(a - b > 0\) we get:
\(b < 0\)
Case 3:if \(a - b < 0, a + b > 0\), then
\(-(a - b) > a + b\)
\(-a + b > a + b\)
\(0 > 2a\)
\(a < 0\)
Also, if we subtract \(a + b > 0\) from \(a - b < 0\) we get:
\(b > 0\)
Case 4:if \(a - b < 0, a + b < 0\), then
\(-(a - b) > -(a + b)\)
\(-a + b > -a - b\)
\(2b > 0\)
\(b > 0\)
Also, if we add \(a + b < 0\) and \(a - b < 0\) we get:
\(a < 0\)
For all of our cases, if \(a < 0\), then \(b > 0\) or if \(a > 0\), then \(b < 0\).
This is the same as \(ab < 0\).
More on this method
here and
here.